Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 61

An ideal gas is contained in a cylinder with a volume of \(5.0 \times\) \(10^{2} \mathrm{~mL}\) at a temperature of \(30 .{ }^{\circ} \mathrm{C}\) and a pressure of 710 . torr. The gas is then compressed to a volume of \(25 \mathrm{~mL}\), and the temperature is raised to \(820 .{ }^{\circ} \mathrm{C}\). What is the new pressure of the gas?

Short Answer

Expert verified
The new pressure of the gas after compression and heating is approximately \(5 \times 10^{4}\) torr.

Step by step solution

01

List the given quantities and convert temperature to Kelvin

Before we begin, it is important to list the given quantities and convert the temperatures to Kelvin since the Ideal Gas Law uses the Kelvin scale. To convert a Celsius temperature to Kelvin, add 273.15. Initial Volume: \(V_1 = 5.0 \times 10^2 \mathrm{~mL}\) Initial Temperature: \(T_1 = 30 .{ }^{\circ} \mathrm{C} = 303.15 \mathrm{K}\) Initial Pressure: \(P_1 = 710 \mathrm{~torr}\) Final Volume: \(V_2 = 25 \mathrm{~mL}\) Final Temperature: \(T_2 = 820 .{ }^{\circ} \mathrm{C} = 1093.15 \mathrm{K}\) We want to find the new pressure \(P_2\).
02

Use Ideal Gas Law for initial and final states

The Ideal Gas Law can be applied to both the initial and final states of the gas. We can construct two equations: Initial state: \(P_1V_1 = nRT_1\) Final state: \(P_2V_2 = nRT_2\)
03

Combine the two equations and solve for the final pressure

Since the amount of gas (n) and the Ideal Gas constant (R) do not change, we can equate the product of initial pressure and volume to the product of final pressure and volume: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) Now, we solve for the final pressure, \(P_2\): \(P_2 = \frac{P_1V_1T_2}{V_2T_1}\)
04

Substitute given values and calculate the final pressure

Now, we can substitute the given values and solve for \(P_2\): \(P_2 = \frac{(710 \mathrm{~torr})(5.0 \times 10^2 \mathrm{~mL})(1093.15 \mathrm{K})}{(25 \mathrm{~mL})(303.15 \mathrm{K})}\) \(P_2 \approx 5 \times 10^{4} \mathrm{~torr}\) Therefore, the new pressure of the gas after compression and heating is approximately \(5 \times 10^{4}\) torr.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Compression
When a gas undergoes compression, its volume decreases while other properties, such as pressure and temperature, may change depending on the conditions of the process. In the case of an ideal gas, a hypothetical gas that perfectly follows the Ideal Gas Law without any intermolecular force influences or volume occupied by the gas molecules, these changes occur in a predictable way.

During compression, if temperature were kept constant (an isothermal process), the pressure of the ideal gas would increase due to Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume. In our problem, the compression is accompanied by a temperature increase, implying that the process is not isothermal. Both the decrease in volume and the increase in temperature act to increase the gas pressure. Therefore, to accurately determine the new pressure following compression, we need to consider both volume change and temperature change, as done through the application of the Combined Gas Law derived from the Ideal Gas Law.
Temperature Conversion
Temperature has a vital role in the behavior of gases and must be measured on an absolute scale for proper scientific calculations. The Kelvin scale is the temperature scale of choice in most scientific work because it is an absolute scale, starting at absolute zero, where all thermal motion ceases.

To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature. This is crucial when using the Ideal Gas Law, as it requires temperature to be in Kelvin. In our exercise, the given temperatures are first converted from Celsius to Kelvin. This avoids potential errors and ensures the accuracy of calculations involved in the Ideal Gas Law, which directly relates pressure, volume, and temperature of an ideal gas.
Pressure-Volume Relationship
The relationship between pressure and volume of a gas is central to understanding gas behavior under varying conditions. According to Boyle's Law, for a given mass of gas at a constant temperature, the volume of the gas is inversely proportional to its pressure. This means that when the volume decreases, the pressure increases proportionally, provided the temperature remains unchanged (isothermal process).

In the provided exercise, the initial and final pressures and volumes of the gas are related through the Ideal Gas Law, which effectively combines Boyle's Law with Charles's Law (the relationship between volume and temperature) and Gay-Lussac's Law (the relationship between pressure and temperature). The use of the Ideal Gas Law accommodates the changes in pressure, volume, and temperature simultaneously, allowing for a precise calculation of the new pressure after the gas has been compressed and the temperature has been raised.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider separate \(1.0\) - \(\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g)\), both at \(1.00\) atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

The total mass that can be lifted by a balloon is given by the difference between the mass of air displaced by the balloon and the mass of the gas inside the balloon. Consider a hot-air balloon that approximates a sphere \(5.00 \mathrm{~m}\) in diameter and contains air heated to \(65^{\circ} \mathrm{C}\). The surrounding air temperature is \(21^{\circ} \mathrm{C}\). The pressure in the balloon is equal to the atmospheric pressure, which is 745 torr. a. What total mass can the balloon lift? Assume that the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol}\). (Hint: Heated air is less dense than cool air.) b. If the balloon is filled with enough helium at \(21^{\circ} \mathrm{C}\) and 745 torr to achieve the same volume as in part a, what total mass can the balloon lift? c. What mass could the hot-air balloon in part a lift if it were on the ground in Denver, Colorado, where a typical atmospheric pressure is 630 . torr?

Which of the following statements is(are) true? For the false statements, correct them. a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules. b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules. c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly. d. As temperature increases, the effect of interparticle interactions on gas behavior is increased. e. At constant \(V\) and \(T\), as gas molecules are added into a container, the number of collisions per unit area increases resulting in a higher pressure. f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and moles of gas.

Consider two separate gas containers at the following conditions: $$ \begin{array}{|ll|} \text { Container A } & \text { Container B } \\ \text { Contents: } \mathrm{SO}_{2}(g) & \text { Contents: unknown gas } \\ \text { Pressure }=P_{\mathrm{A}} & \text { Pressure }=P_{\mathrm{B}} \\ \text { Moles of gas }=1.0 \mathrm{~mol} & \text { Moles of gas }=2.0 \mathrm{~mol} \\ \text { Volume }=1.0 \mathrm{~L} & \text { Volume }=2.0 \mathrm{~L} \\ \text { Temperature }=7^{\circ} \mathrm{C} & \text { Temperature }=287^{\circ} \mathrm{C} \\ \hline \end{array} $$ How is the pressure in container \(\mathrm{B}\) related to the pressure in container \(\mathrm{A}\) ?

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a 10.000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 115 .
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free